A solid sphere is released from the top of a ramp that is at a height h 1 = 2.30

Question

A solid sphere is released from the top of a ramp that is at a height

h1 = 2.30 m.

It goes down the ramp, the bottom of which is at a height of

h2 = 1.73 m

above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.16 m.

A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It goes down the ramp, the bottom of which is at a height of h2 = 1.73 m above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.16 m. (a) Through what horizontal distance d does the ball travel before landing? Consider the motion along the ramp first. What is the speed of the ball at the bottom of the ramp? Now consider the motion of the ball after it leaves the ramp. What is the path of the ball once it leaves the ramp? m (b) How many revolutions does the ball make during its fall?

Explanation / Answer

a)
let v is the speed at height h2,

Apply,loss of PE = gain in KE

m*g*(h1-h2) = 0.5*m*v^2 + 0.5*I*w^2

m*g*(h1-h2) = 0.5*m*v^2 + 0.5*(2/5)*m*r^2*w^2

m*g*(h1-h2) = 0.5*m*v^2 + 0.2*m*v^2

m*g*(h1-h2) = 0.7*m*v^2

v = sqrt(g*(h1-h2)/0.7)

= sqrt(9.8*(2.3-1.73)/0.7)

= 2.82 m/s

let t is the time taken,

h2 = 0.5*g*t^2

t = sqrt(2*h2/g)

= sqrt(2*1.73/9.8)

= 0.594 s

d = v*t

= 0.594*2.82
= 1.68 m <<<<<<<<--------Answer

b) w = v/r

= 2.82/0.08

= 35.25 rad/s

theta = w*t

= 35.25*0.594

= 20.94 rad

= 20.94/(2*pi)

= 3.33 revolutions <<<<<<<<--------Answer

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